Solution to:
Word Sums

It is given that the digits 1 and 6 are the most frequently used, so we first look how often each letter occurs in the puzzle:

M, V, and P occur once;

T occurs twice;

A, R, and E occur three times;

S occurs four times;

N and U occur five times.

So either N=1 and U=6, or N=6 and U=1.
Since even 9999 + 99999 + 999999 + 999999 is less than 6000000, N cannot be 6.
So N=1 and U=6.

We now have:

mars
ve16s
6ra16s
sat6r1
------- +
1ept61e

R+6+6+R (in the 'tens' column) is even, but since its sum ends on a 1, 1 must have been carried from the 'units' column.
We can also see that R=4 or R=9.
The fact that 1 is carried from the 'units' to the 'tens' column gives the following possible values for S:

S=3, so S+S+S+1=10, and therefore E=0;

S=4, so S+S+S+1=13, and therefore E=3;

S=5, so S+S+S+1=16, and therefore E=6: this is not possible because U=6.

From the previous step, we know that R=4 or R=9.
We first look at the case S=3.
If R would be 4, a value of 2 would be carried from the 'tens' to the 'hundreds' column.
So the sum of 2+A+1+1+6 must end on a 6, which gives A=6.
This is not possible because U=6.
So, if S=3, then R must be 9 and a value of 3 is carried from the 'tens' to the 'hundreds' column.
So the sum of 3+A+1+1+6 must end on a 6, which gives A=5.

We now look at the case S=4.
In this case, R can only be 9.
Then a value of 3 is carried from the 'tens' to the 'hundreds' column, which gives A=5.

Therefore, regardless of the values of S and E, we know that R=9 and A=5.
Depending on the values of S and E, we now have:

In either case, a value of 1 is carried from the 'hundreds' to the 'thousands' column.
In the case that S=3 and E=0, the sum of 1+M+0+5+T must end on T, which gives M=4.
In the case that S=4 and E=3, the sum of 1+M+3+5+T must end on T, which gives M=1.
This, however, is not possible because U=1.
From this, it follows that S cannot be 4, so we now know that S=3, E=0, and M=4.

We now have:

4593
v0163
695163
35t691
------- +
10pt610

The only remaining digits are now 2, 7, and 8.
Since a value of 1 is carried from the 'thousands' to the 'ten thousands' column, it follows that V=2 and P=7.
For T, only the value 8 remains.